Question

A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0\(^\circ\) below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 3.00 m farther down the beam and perpendicular to it (\(\textbf{Fig. E11.20}\)). The center of gravity of the beam is 2.00 m down the beam from the pivot. Lighting equipment exerts a 5.00-kN downward force on the lower left end of the beam. Find the tension \(T\) in the cable and the horizontal and vertical components of the force exerted on the beam by the pivot. Start by sketching a free-body diagram of the beam.

Short answer

Tension in the cable is 7.35 kN; horizontal force at the pivot is 3.11 kN; vertical force is approximately -0.26 kN.

Step-by-step solution

Sketch the Free-Body Diagram

Draw the beam where it makes a 25-degree angle above from horizontal. Indicate the pivot at the upper right end, draw the cable perpendicular to the beam at 3.00 m from the pivot, and note the center of gravity at 2.00 m from the pivot. Show the 1.40 kN force acting downward at the center of gravity, the 5.00 kN force at the lower left end of the beam, and label the tension in the cable and the forces at the pivot (both horizontal and vertical).

Define the Coordinate System and Forces

Let the x-axis be horizontal and the y-axis be vertical. Define the forces: - Tension in the cable as T. - Weight of the beam (1.40 kN) acting vertically downward at 25 degrees from the horizontal at 2.00 m from the pivot. - The force due to the lighting equipment (5.00 kN) acting vertically downward at the end of the beam. - Reaction forces at the pivot: horizontal (F_h) and vertical (F_v).

Calculate Moments Around the Pivot

Using the pivot as the reference point, the sum of clockwise and counterclockwise moments must be zero:\[ (1.40 \times 2)\cos(25) + 5.00 \times 4.5\cos(25) = T \times 3 \]Simplifying gives:\[ 2.54 \text{ kNm} + 19.5 \text{ kNm} = T \times 3 \]Solve for T (in kN).

Calculate Tension in the Cable

Solving for T in the equation:\[ T \times 3 = 22.04 \]Gives:\[ T = \frac{22.04}{3} \approx 7.35 \text{ kN} \]

Resolve Forces in the Horizontal and Vertical Directions

Using equilibrium conditions for forces, resolve:- Horizontal: the sum of forces horizontal = 0, so\[ F_h = T \sin(25) \]- Vertical: the sum of forces vertical = 0, so\[ F_v = 1.40 + 5.00 - T \cos(25) \]

Calculate Horizontal Force (F_h) at the Pivot

Using the value of T from Step 4:\[ F_h = 7.35 \sin(25) \approx 3.11 \text{ kN} \]

Calculate Vertical Force (F_v) at the Pivot

Continuing from Step 5, substitute the value of T:\[ F_v = 1.40 + 5.00 - 7.35 \cos(25) \approx 1.40 + 5.00 - 6.66 \approx -0.26 \text{ kN} \] (The negative sign indicates downward direction.)

Key concepts

Free-Body Diagram

A free-body diagram is an essential tool in physics that helps to visualize forces acting on an object. When you sketch a free-body diagram, you should begin by illustrating the object, in this case, the beam. The beam is tilted at a 25-degree angle below the horizontal. It's important to accurately represent the pivot, located at the beam's upper right end, and the cable, 3.00 meters down the beam from the pivot and perpendicular to the beam.

Include forces acting on the beam: the beam's weight of 1.40 kN acts downward at its center of gravity, positioned 2.00 meters from the pivot. There's also a 5.00 kN force from lighting equipment acting at the beam's lower end. Don't forget to illustrate the tension in the cable and the forces at the pivot, which consist of both horizontal and vertical components. Properly labeling these forces is crucial to avoiding confusion later on.

Tension Calculation

Calculating the tension in the cable involves understanding the interplay of moments around the pivot point. To find the tension, apply the principle of moments, which states that for an object in equilibrium, the sum of clockwise moments around any point must be equal to the sum of counterclockwise moments.

Start by taking the pivot as the point of reference. You calculate the moments due to the weight of the beam and the lighting equipment, both acting as counterclockwise moments. Then, find the tension's moment, which must balance those moments. The equation you use is:\[ (1.40 \times 2)\cos(25) + 5.00 \times 4.5\cos(25) = T \times 3 \]

Upon solving, it gives:\[ T = \frac{22.04}{3} \approx 7.35 \text{kN} \]
This value represents the tension required in the cable to keep the beam in equilibrium.

Force Components

When resolving forces into components, it helps to picture how these forces act in different directions. The beam has both horizontal (x-axis) and vertical (y-axis) components to consider. For the horizontal force components, tension acts at an angle, creating horizontal force at the pivot:\[ F_h = T \sin(25) \]
By inserting the tension value, you find:\[ F_h = 7.35 \sin(25) \approx 3.11 \text{kN} \]

Moving to the vertical components, the balance of vertical forces must sum to zero for equilibrium. All vertical acting forces, including beam weight, lighting, and tension, affect the vertical force:\[ F_v = 1.40 + 5.00 - T \cos(25) \]
Inserting the tension gives the vertical force at the pivot:\[ F_v = 1.40 + 5.00 - 6.66 \approx -0.26 \text{kN} \]
The negative sign indicates that the vertical force direction is downward.

Equilibrium Conditions

For the beam to remain in equilibrium, all forces and moments acting on it need to satisfy the conditions of static equilibrium. This means the sum of all horizontal forces, vertical forces, and moments must equal zero.

Breaking it down:

• The sum of horizontal forces: This condition ensures no net force is pushing the beam sideways, allowing it to stay in place horizontally. Here, the tension’s horizontal component is counteracted by the pivot's force.
• The sum of vertical forces: For vertical balance, the downward forces induced by the beam's weight and the lighting must balance upward forces, including the pivot's force.
• Moments about the pivot: As previously discussed, this condition ensures the beam doesn’t rotate. The counterbalance between forces prevents turning around the pivot.

When each of these conditions is satisfied, the beam is at equilibrium. They collectively ascertain that the beam remains steady, preventing any unwanted movement or rotation.